"def flatten_dictionary(dictionary):
\# return a flattened dictionary - int/string/another dictionary values
\# if the key is empty, exclude from the output
\# concat using a "." btwn them
\# add to res which is { "key.a.b.etc": "value" }
\# iterate through the key value pairs
\# while there is a key value pair in the value
\# continue going through that, until the value is an int/string
flatDic = {}
flatDicHelper("", dictionary, flatDic)
print(flatDic)
return flatDic
def flatDicHelper(initialKey"
Anonymous Owl - "def flatten_dictionary(dictionary):
\# return a flattened dictionary - int/string/another dictionary values
\# if the key is empty, exclude from the output
\# concat using a "." btwn them
\# add to res which is { "key.a.b.etc": "value" }
\# iterate through the key value pairs
\# while there is a key value pair in the value
\# continue going through that, until the value is an int/string
flatDic = {}
flatDicHelper("", dictionary, flatDic)
print(flatDic)
return flatDic
def flatDicHelper(initialKey"See full answer
"Sorting is a technique to arrange data in either increasing order or decreasing order, and, the function that implements this functionality is called sort function"
Farhan -. - "Sorting is a technique to arrange data in either increasing order or decreasing order, and, the function that implements this functionality is called sort function"See full answer
"with cte as
(select *,
row_number() over(order by score desc) as rn
from players)
select player_name, score, rn as ranking
from cte
where rn= 4 or rn =6 or rn =11
`"
Gowtami K. - "with cte as
(select *,
row_number() over(order by score desc) as rn
from players)
select player_name, score, rn as ranking
from cte
where rn= 4 or rn =6 or rn =11
`"See full answer
"Construct a min-heap either inplace, or by making a copy of the array and then applying heapify on that copy. This is done in O(n) time.
Maintain two zero-initialised variables - sum and count.
Keep popping off the heap while sum < k, and update count.
In the worst case you will have to do n pops, and each pop is O(log n), so the algorithm would take O(n log n) in total. Space complexity depends on whether you're allowed to modify inplace or not, so either O(1) or O(n) respectively."
Anonymous Wolf - "Construct a min-heap either inplace, or by making a copy of the array and then applying heapify on that copy. This is done in O(n) time.
Maintain two zero-initialised variables - sum and count.
Keep popping off the heap while sum < k, and update count.
In the worst case you will have to do n pops, and each pop is O(log n), so the algorithm would take O(n log n) in total. Space complexity depends on whether you're allowed to modify inplace or not, so either O(1) or O(n) respectively."See full answer
"WITH CTE AS (
SELECT *, ROWNUMBER()OVER(PARTITION BY utxoid ORDER BY transactionid) AS trxrk
FROM transactions
JOIN transaction_inputs
USING (transaction_id)
JOIN utxo
USING (utxo_id)
)
SELECT transaction_id AS InvalidTransactionId
FROM CTE
WHERE sender!=address OR trx_rk > 1
`"
E L. - "WITH CTE AS (
SELECT *, ROWNUMBER()OVER(PARTITION BY utxoid ORDER BY transactionid) AS trxrk
FROM transactions
JOIN transaction_inputs
USING (transaction_id)
JOIN utxo
USING (utxo_id)
)
SELECT transaction_id AS InvalidTransactionId
FROM CTE
WHERE sender!=address OR trx_rk > 1
`"See full answer
"SELECT
e1.empid AS manageremployee_id,
e1.empname AS managername,
COUNT(e2.empid) AS numberofdirectreports
FROM employees AS e1
INNER JOIN employees AS e2
ON e2.managerid = e1.empid
GROUP BY e1.emp_id
HAVING COUNT(e2.emp_id) >= 2
ORDER BY numberofdirectreports DESC, managername ASC
`"
Alvin P. - "SELECT
e1.empid AS manageremployee_id,
e1.empname AS managername,
COUNT(e2.empid) AS numberofdirectreports
FROM employees AS e1
INNER JOIN employees AS e2
ON e2.managerid = e1.empid
GROUP BY e1.emp_id
HAVING COUNT(e2.emp_id) >= 2
ORDER BY numberofdirectreports DESC, managername ASC
`"See full answer
"// Helper function to calculate the Euclidean distance between two points
function distance(p1, p2) {
return Math.sqrt(Math.pow(p1[0] - p2[0], 2) + Math.pow(p1[1] - p2[1], 2));
}
// A helper function to find the closest pair in a given set of points within the strip
function closestPairInStrip(strip, d) {
let minDist = d; // Start with the current minimum distance
strip.sort((a, b) => a[1] - b[1]); // Sort the strip by y-coordinate
for (let i = 0; i < strip.length; i++) {
"
Vishnu V. - "// Helper function to calculate the Euclidean distance between two points
function distance(p1, p2) {
return Math.sqrt(Math.pow(p1[0] - p2[0], 2) + Math.pow(p1[1] - p2[1], 2));
}
// A helper function to find the closest pair in a given set of points within the strip
function closestPairInStrip(strip, d) {
let minDist = d; // Start with the current minimum distance
strip.sort((a, b) => a[1] - b[1]); // Sort the strip by y-coordinate
for (let i = 0; i < strip.length; i++) {
"See full answer
"Determine the requirements
Perform basic checks on the frontend before submitting to the server
Length
Check for invalid or required characters
Client-side validation messaging
Perform more advanced checks on the server if required
Does the password include the user's name, etc
Handle server-side validation messaging"
Casey C. - "Determine the requirements
Perform basic checks on the frontend before submitting to the server
Length
Check for invalid or required characters
Client-side validation messaging
Perform more advanced checks on the server if required
Does the password include the user's name, etc
Handle server-side validation messaging"See full answer
"Bitshift the number to the right and keep track of the 1's you encounter. If you bitshift it completely and only encounter one 1, it is a power of two."
Nils G. - "Bitshift the number to the right and keep track of the 1's you encounter. If you bitshift it completely and only encounter one 1, it is a power of two."See full answer
"def findAlibaba(countOfRooms, strategy):
#countofrooms: num rooms
#listRooms rooms to look for alibabba
possiblePlaces = []
#initialize rooms
for i in range(countOfRooms):
possiblePlaces.append(True)
for i in range(len(strategy)):
roomToCheck = strategy[i]
#Room is marked as unavailable
possiblePlaces[roomToCheck] = False
#Next day calculatins
nextDayPlaces = []
for j in range(countOfRooms):
nextDayPla"
JOBHUNTER - "def findAlibaba(countOfRooms, strategy):
#countofrooms: num rooms
#listRooms rooms to look for alibabba
possiblePlaces = []
#initialize rooms
for i in range(countOfRooms):
possiblePlaces.append(True)
for i in range(len(strategy)):
roomToCheck = strategy[i]
#Room is marked as unavailable
possiblePlaces[roomToCheck] = False
#Next day calculatins
nextDayPlaces = []
for j in range(countOfRooms):
nextDayPla"See full answer
"SELECT COUNT(DISTINCT o.customerid) AS customers, d.departmentname
FROM orders o
INNER JOIN departments d
ON d.departmentid = o.departmentid
WHERE d.departmentname IN ('Electronics','Fashion') AND o.orderdate BETWEEN '2022-01-01' AND '2022-12-31'
GROUP BY d.department_name;
`"
Derrick M. - "SELECT COUNT(DISTINCT o.customerid) AS customers, d.departmentname
FROM orders o
INNER JOIN departments d
ON d.departmentid = o.departmentid
WHERE d.departmentname IN ('Electronics','Fashion') AND o.orderdate BETWEEN '2022-01-01' AND '2022-12-31'
GROUP BY d.department_name;
`"See full answer